A set $A\subseteq \mathbb{R^n}$ is open iff $\mathbb{R^n}-A$ is closed.

A set $A\subseteq \mathbb{R^n}$ is closed iff $\mathbb{R^n}-A$ is open.

Proof. Suppose that $A$ is open. We must show that for every $x\notin \mathbb{R^n}-A$ there is a nbhd $N$ with $N\cap (\mathbb{R^n}-A)=\varnothing$. Let $x\notin \mathbb{R^n}-A\implies x\in A$. As $A$ is open, there is an open nbhd $N$ of $x$ such that $x\in N \subseteq A$. This implies that $N\cap (\mathbb{R^n}-A)=\varnothing$.

1) $(\implies)$ You are almost there, assume $A$ is an open set. Let $x\in A$, then we can find an open set $O$ such that $x\in O \subset A $ which implies $O \cap A^c = \phi$ since $A \cap A^c = \phi.$